send edited codeMain 2public class Main2 { public static voi.pdf
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send edited code Main 2 public class Main2 { public static void main(String[] args) { EZ.initialize(512, 512); EZ.addImage(\"metropolis.jpg\", 256, 256); Hero myHero = new Hero(100,390); //HeroPlus mySecondHero = new HeroPlus(100,390); while(true){ myHero.processStates(); //mySecondHero.processStates(); EZ.refreshScreen(); } } } Hero Plus public class HeroPlus { EZImage heroStand; EZImage heroJump; EZImage heroLand; EZImage heroHover; EZImage heroPunch; // PUNCH int posx = 0; int posy = 0; int heroState = STAND; static final int STAND = 1; static final int JUMP = 2; static final int LAND = 3; static final int HOVER = 4; static final int PUNCH = 5; static final int JUMPHEIGHT = 100; static final int PUNCHDISTANCE = 50; // PUNCH int punchCounter = 0; // PUNCH int jumpCounter = 0; HeroPlus(int x, int y) { posx = x; posy = y; heroStand = EZ.addImage(\"superstand.png\", posx, posy); heroLand = EZ.addImage(\"superland.png\", posx, posy); heroJump = EZ.addImage(\"superjump.png\", posx, posy); heroHover = EZ.addImage(\"superhover.png\", posx, posy); heroPunch = EZ.addImage(\"superpunch.png\", posx, posy); // PUNCH hideHero(); heroStand.show(); } void hideHero() { heroStand.hide(); heroLand.hide(); heroJump.hide(); heroHover.hide(); heroPunch.hide(); // PUNCH } void positionHero(int x, int y) { heroStand.translateTo(x, y); heroJump.translateTo(x, y); heroLand.translateTo(x, y); heroHover.translateTo(x, y); heroPunch.translateTo(x, y); // PUNCH } void processStates() { switch (heroState) { case STAND: if (EZInteraction.wasKeyPressed(\'j\')) { heroState = JUMP; jumpCounter = 0; hideHero(); heroJump.show(); } break; case JUMP: jumpCounter++; if (EZInteraction.wasKeyPressed(\'p\')) { punchCounter = 0; hideHero(); heroPunch.show(); heroState = PUNCH; } if (jumpCounter > JUMPHEIGHT) { heroState = LAND; hideHero(); heroLand.show(); } else { posy -= 2; positionHero(posx, posy); if (EZInteraction.wasKeyPressed(\'h\')) { heroState = HOVER; hideHero(); heroHover.show(); } } break; case PUNCH: // PUNCH punchCounter++; posx += 2; positionHero(posx, posy); if (punchCounter > PUNCHDISTANCE) { heroState = LAND; hideHero(); heroLand.show(); } break; case LAND: jumpCounter--; if (jumpCounter <= 0) { heroState = STAND; hideHero(); heroStand.show(); } else { posy += 2; positionHero(posx, posy); if (EZInteraction.wasKeyPressed(\'h\')) { heroState = HOVER; hideHero(); heroHover.show(); } } break; case HOVER: if (EZInteraction.wasKeyPressed(\'l\')) { heroState = LAND; hideHero(); heroLand.show(); } break; } } } Solution public class Main2 { public static void main(String[] args) { EZ.initialize(512, 512); EZ.addImage(\"metropolis.jpg\", 256, 256); // Hero myHero = new Hero(100,390); HeroPlus mySecondHero = new HeroPlus(100,390); while(true){ //myHero.processStates(); mySecondHero.processStates(); EZ.refreshScreen(); } } } class HeroPlus { EZImage heroStand; EZImage heroJump; EZImage heroLand; EZImage heroHover; EZImage heroPunch; // PUNCH int posx = 0; int posy = 0; int her.
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Main 2
public class Main2 {
public static void main(String[] args) {
EZ.initialize(512, 512);
EZ.addImage("metropolis.jpg", 256, 256);
Hero myHero = new Hero(100,390);
//HeroPlus mySecondHero = new HeroPlus(100,390);
while(true){
myHero.processStates();
//mySecondHero.processStates();
EZ.refreshScreen();
}
}
}
Hero Plus
public class HeroPlus {
EZImage heroStand;
EZImage heroJump;
EZImage heroLand;
EZImage heroHover;
EZImage heroPunch; // PUNCH
int posx = 0;
int posy = 0;
int heroState = STAND;
static final int STAND = 1;
static final int JUMP = 2;
static final int LAND = 3;
static final int HOVER = 4;
static final int PUNCH = 5;
static final int JUMPHEIGHT = 100;](https://image.slidesharecdn.com/sendeditedcodemain2publicclassmain2publicstaticvoi-230708024420-7e9a1100/85/send-edited-codeMain-2public-class-Main2-public-static-voi-pdf-1-320.jpg)
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This is Java,
I am currently stumped on how to add a scoreboard for my game that I am making. I have
inclued my code and classes so far. Any help with a working scoreboard would be greatly
apperiacted.
Game.java
import javax.swing.JFrame;
public class Game {
public static void main(String[] args)
{
// create the frame
JFrame myFrame = new JFrame(\"Platformer\");
// set up the close operation
myFrame.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
// create panel
Mainpanel myPanel = new Mainpanel();
// add panel
myFrame.getContentPane().add(myPanel);
// pack
myFrame.pack();
// set visibility to true
myFrame.setVisible(true);
}
}
Mainpanel.java
import java.awt.Dimension;
import java.awt.Graphics;
import java.awt.Graphics2D;
import java.awt.event.KeyEvent;
import java.awt.event.KeyListener;
import java.util.ArrayList;
import java.util.Random;
import java.awt.Color;
import java.awt.event.ActionEvent;
import java.awt.event.ActionListener;
import javax.swing.ImageIcon;
import javax.swing.JLabel;
import javax.swing.JPanel;
import javax.swing.Timer;
public class Mainpanel extends JPanel implements KeyListener, ActionListener{
private final int boardWidth =1000;
private final int boardHeight =1000;
int x = 0;
int y = 0;
int i= 0;
int goldCount=11;
int score = 0;
ImageIcon myIcon = new ImageIcon(\"./src/TreasureChest.png\");
Timer mainTimer;
player player1;
player player2;
static ArrayList treasure = new ArrayList();
Random rand = new Random();
public String ScoreCount = \"Score: \" + score;
public Mainpanel()
{
setPreferredSize(new Dimension(boardWidth,boardHeight));
addKeyListener(this);
setFocusable(true);
player1= new player (100,100);
player2= new player (200,200);
addKeyListener(new move(player1));
addKeyListener(new move(player2));
mainTimer = new Timer(10,this);
mainTimer.start();
startGame();
}
JLabel scoreLabel = new JLabel(\"Score: 0\");
public void paintComponent(Graphics page)
{
super.paintComponent(page);
Graphics2D g2d =(Graphics2D) page;
player1.draw(g2d);
player2.draw(g2d);
g2d.
g2d.setColor(new Color(128, 128, 128));
g2d.fillRect(0, 0, 50, 1000);
g2d.setColor(new Color(128, 128, 128));
g2d.fillRect(950, 0, 50, 1000);
g2d.setColor(new Color(128, 128, 128));
g2d.fillRect(50, 0, 900, 50);
g2d.setColor(new Color(128, 128, 128));;
g2d.fillRect(50, 950, 900, 50);
for (int i=0 ; i < treasure.size(); i++){
Gold tempGold = treasure.get(i);
tempGold.draw(g2d);
}
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public void actionPerformed (ActionEvent arg0){
player1.update();
repaint();
}
@Override
public void keyPressed(KeyEvent arg0) {
// TODO Auto-generated method stub
}
@Override
public void keyReleased(KeyEvent arg0) {
// TODO Auto-generated method stub
}
@Override
public void keyTyped(KeyEvent arg0) {
// TODO Auto-generated method stub
}
public void addGold(Gold g){
treasure.add(g);
}
public static void removeGold (Gold g) {
treasure.remove(g);
}
public static ArrayList getGoldList() {
return treasure;
}
public void startGame() {
for (int i=0; i < goldCount; i++){
addGold(new Gold(rand.
ObjectiveCreate a graphical game of minesweeper IN JAVA. The boar.pdf
Objective:
Create a graphical game of minesweeper IN JAVA. The board should consist of 10x10 buttons.
Of the 100 spaces there should be at least 20 randomly placed mines. If the button is clicked and
it is not a mine then it clears itself. If a space has been cleared then it should indicate how many
of its eight neighbors are mines. If a space is clicked and it is a mine then the game is over and
the player is asked if they want to play again. Finally, if all the non-mine spaces have been
clicked then the player is prompted that they won.
Solution
2 files need to be made.
First Game.Java which will have the gaming logic and User Interface and Main.Java which will
have have the main method and object of Game.java Number of rows and cols is customizable so
user can change that as and when needed.
import java.awt.*;
import java.awt.Dimension;
import javax.swing.*;
import java.awt.event.*;
import java.util.*;
public class Game extends JFrame implements ActionListener, ContainerListener {
int fw, fh, blockr, blockc, var1, var2, num_of_mine, detectedmine = 0, savedlevel = 1,
savedblockr, savedblockc, savednum_of_mine = 10;
int[] r = {-1, -1, -1, 0, 1, 1, 1, 0};
int[] c = {-1, 0, 1, 1, 1, 0, -1, -1};
JButton[][] blocks;
int[][] countmine;
int[][] colour;
ImageIcon[] ic = new ImageIcon[14];
JPanel panelb = new JPanel();
JPanel panelmt = new JPanel();
JTextField tf_mine, tf_time;
JButton reset = new JButton(\"\");
Random ranr = new Random();
Random ranc = new Random();
boolean check = true, starttime = false;
Point framelocation;
Stopwatch sw;
MouseHendeler mh;
Point p;
Game() {
super(\"Game\");
setLocation(400, 300);
setic();
setpanel(1, 0, 0, 0);
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sw = new Stopwatch();
reset.addActionListener(new ActionListener() {
public void actionPerformed(ActionEvent ae) {
try {
sw.stop();
setpanel(savedlevel, savedblockr, savedblockc, savednum_of_mine);
} catch (Exception ex) {
setpanel(savedlevel, savedblockr, savedblockc, savednum_of_mine);
}
reset();
}
});
setDefaultCloseOperation(EXIT_ON_CLOSE);
show();
}
public void reset() {
check = true;
starttime = false;
for (int i = 0; i < blockr; i++) {
for (int j = 0; j < blockc; j++) {
colour[i][j] = \'w\';
}
}
}
public void setpanel(int level, int setr, int setc, int setm) {
if (level == 1) {
fw = 200;
fh = 300;
blockr = 10;
blockc = 10;
num_of_mine = 10;
} else if (level == 2) {
fw = 320;
fh = 416;
blockr = 16;
blockc = 16;
num_of_mine = 70;
} else if (level == 3) {
fw = 400;
fh = 520;
blockr = 20;
blockc = 20;
num_of_mine = 150;
} else if (level == 4) {
fw = (20 * setc);
fh = (24 * setr);
blockr = setr;
blockc = setc;
num_of_mine = setm;
}
savedblockr = blockr;
savedblockc = blockc;
savednum_of_mine = num_of_mine;
setSize(fw, fh);
setResizable(false);
detectedmine = num_of_mine;
p = this.getLocation();
blocks = new JButton[blockr][blockc];
countmine = new int[blockr][blockc];
colour = new int[blockr][blockc];
mh = new MouseHendeler();
getContentPane().removeAll();
panelb.removeAll();
tf_mine = new JTe.
What two kinds of secondary bonds hold polymer chains together Which.pdf
What two kinds of secondary bonds hold polymer chains together? Which is stronger?
Solution
there are three kinds of secondary forces in polymers
1. Polar
2. Hydrogen Bond
3. Non Polar
Out of these three polar bond is more strong because these bond are formed between a
electronegative atom such as oxygen, fluorine and positive atoms like hydrogen and carbon. This
bond has characteristics of Ionic bond and hence it is stronger..
When hiking up most Stratovolcanoes, which rock composition(s) will .pdf
When hiking up most Stratovolcanoes, which rock composition(s) will you see from its base to
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the bottom, intermediate in the middle, and mafic at the top c. Intermediate from bottom to top d.
Felsic from bottom to top
Solution
option (a) is correct
A . Mafic at the bottom ,
intermediate in the middle ,
and felsic at the top ..
What image of society is held by the three major theoretical approac.pdf
What image of society is held by the three major theoretical approaches? What core questions
does each approach ask? Which of the three approaches strikes you as the most useful? Why?
Solution
The three major theoretical approaches:-
1) Symbolic interactionism
2) Conflict Theory
3) Functionalism
Explanation:-
1) Symbolic interactionism:- emphasises on the language and symbols that help us to give
meaning to the experiences in our life. It is basically slices of society. It is distinctive approach to
the study of human life and human conduct.
2) Conflict Theory:- This theory argues that individuals and groups i.e., social classes within
society have different amounts of material and non-material resources (such as the wealthy vs.
the poor) and that the more powerful groups use their dominance and power in order to exploit
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3) Functionalism:-This theory focuses how everyone take a part in society, how is the behaviour
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This is Java,I am currently stumped on how to add a scoreboard for.pdf
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apperiacted.
Game.java
import javax.swing.JFrame;
public class Game {
public static void main(String[] args)
{
// create the frame
JFrame myFrame = new JFrame(\"Platformer\");
// set up the close operation
myFrame.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
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myFrame.setVisible(true);
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Mainpanel.java
import java.awt.Dimension;
import java.awt.Graphics;
import java.awt.Graphics2D;
import java.awt.event.KeyEvent;
import java.awt.event.KeyListener;
import java.util.ArrayList;
import java.util.Random;
import java.awt.Color;
import java.awt.event.ActionEvent;
import java.awt.event.ActionListener;
import javax.swing.ImageIcon;
import javax.swing.JLabel;
import javax.swing.JPanel;
import javax.swing.Timer;
public class Mainpanel extends JPanel implements KeyListener, ActionListener{
private final int boardWidth =1000;
private final int boardHeight =1000;
int x = 0;
int y = 0;
int i= 0;
int goldCount=11;
int score = 0;
ImageIcon myIcon = new ImageIcon(\"./src/TreasureChest.png\");
Timer mainTimer;
player player1;
player player2;
static ArrayList treasure = new ArrayList();
Random rand = new Random();
public String ScoreCount = \"Score: \" + score;
public Mainpanel()
{
setPreferredSize(new Dimension(boardWidth,boardHeight));
addKeyListener(this);
setFocusable(true);
player1= new player (100,100);
player2= new player (200,200);
addKeyListener(new move(player1));
addKeyListener(new move(player2));
mainTimer = new Timer(10,this);
mainTimer.start();
startGame();
}
JLabel scoreLabel = new JLabel(\"Score: 0\");
public void paintComponent(Graphics page)
{
super.paintComponent(page);
Graphics2D g2d =(Graphics2D) page;
player1.draw(g2d);
player2.draw(g2d);
g2d.
g2d.setColor(new Color(128, 128, 128));
g2d.fillRect(0, 0, 50, 1000);
g2d.setColor(new Color(128, 128, 128));
g2d.fillRect(950, 0, 50, 1000);
g2d.setColor(new Color(128, 128, 128));
g2d.fillRect(50, 0, 900, 50);
g2d.setColor(new Color(128, 128, 128));;
g2d.fillRect(50, 950, 900, 50);
for (int i=0 ; i < treasure.size(); i++){
Gold tempGold = treasure.get(i);
tempGold.draw(g2d);
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public void actionPerformed (ActionEvent arg0){
player1.update();
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@Override
public void keyPressed(KeyEvent arg0) {
// TODO Auto-generated method stub
}
@Override
public void keyReleased(KeyEvent arg0) {
// TODO Auto-generated method stub
}
@Override
public void keyTyped(KeyEvent arg0) {
// TODO Auto-generated method stub
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treasure.add(g);
}
public static void removeGold (Gold g) {
treasure.remove(g);
}
public static ArrayList getGoldList() {
return treasure;
}
public void startGame() {
for (int i=0; i < goldCount; i++){
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ObjectiveCreate a graphical game of minesweeper IN JAVA. The boar.pdf
Objective:
Create a graphical game of minesweeper IN JAVA. The board should consist of 10x10 buttons.
Of the 100 spaces there should be at least 20 randomly placed mines. If the button is clicked and
it is not a mine then it clears itself. If a space has been cleared then it should indicate how many
of its eight neighbors are mines. If a space is clicked and it is a mine then the game is over and
the player is asked if they want to play again. Finally, if all the non-mine spaces have been
clicked then the player is prompted that they won.
Solution
2 files need to be made.
First Game.Java which will have the gaming logic and User Interface and Main.Java which will
have have the main method and object of Game.java Number of rows and cols is customizable so
user can change that as and when needed.
import java.awt.*;
import java.awt.Dimension;
import javax.swing.*;
import java.awt.event.*;
import java.util.*;
public class Game extends JFrame implements ActionListener, ContainerListener {
int fw, fh, blockr, blockc, var1, var2, num_of_mine, detectedmine = 0, savedlevel = 1,
savedblockr, savedblockc, savednum_of_mine = 10;
int[] r = {-1, -1, -1, 0, 1, 1, 1, 0};
int[] c = {-1, 0, 1, 1, 1, 0, -1, -1};
JButton[][] blocks;
int[][] countmine;
int[][] colour;
ImageIcon[] ic = new ImageIcon[14];
JPanel panelb = new JPanel();
JPanel panelmt = new JPanel();
JTextField tf_mine, tf_time;
JButton reset = new JButton(\"\");
Random ranr = new Random();
Random ranc = new Random();
boolean check = true, starttime = false;
Point framelocation;
Stopwatch sw;
MouseHendeler mh;
Point p;
Game() {
super(\"Game\");
setLocation(400, 300);
setic();
setpanel(1, 0, 0, 0);
setmanue();
sw = new Stopwatch();
reset.addActionListener(new ActionListener() {
public void actionPerformed(ActionEvent ae) {
try {
sw.stop();
setpanel(savedlevel, savedblockr, savedblockc, savednum_of_mine);
} catch (Exception ex) {
setpanel(savedlevel, savedblockr, savedblockc, savednum_of_mine);
}
reset();
}
});
setDefaultCloseOperation(EXIT_ON_CLOSE);
show();
}
public void reset() {
check = true;
starttime = false;
for (int i = 0; i < blockr; i++) {
for (int j = 0; j < blockc; j++) {
colour[i][j] = \'w\';
}
}
}
public void setpanel(int level, int setr, int setc, int setm) {
if (level == 1) {
fw = 200;
fh = 300;
blockr = 10;
blockc = 10;
num_of_mine = 10;
} else if (level == 2) {
fw = 320;
fh = 416;
blockr = 16;
blockc = 16;
num_of_mine = 70;
} else if (level == 3) {
fw = 400;
fh = 520;
blockr = 20;
blockc = 20;
num_of_mine = 150;
} else if (level == 4) {
fw = (20 * setc);
fh = (24 * setr);
blockr = setr;
blockc = setc;
num_of_mine = setm;
}
savedblockr = blockr;
savedblockc = blockc;
savednum_of_mine = num_of_mine;
setSize(fw, fh);
setResizable(false);
detectedmine = num_of_mine;
p = this.getLocation();
blocks = new JButton[blockr][blockc];
countmine = new int[blockr][blockc];
colour = new int[blockr][blockc];
mh = new MouseHendeler();
getContentPane().removeAll();
panelb.removeAll();
tf_mine = new JTe.
What two kinds of secondary bonds hold polymer chains together Which.pdf
What two kinds of secondary bonds hold polymer chains together? Which is stronger?
Solution
there are three kinds of secondary forces in polymers
1. Polar
2. Hydrogen Bond
3. Non Polar
Out of these three polar bond is more strong because these bond are formed between a
electronegative atom such as oxygen, fluorine and positive atoms like hydrogen and carbon. This
bond has characteristics of Ionic bond and hence it is stronger..
When hiking up most Stratovolcanoes, which rock composition(s) will .pdf
When hiking up most Stratovolcanoes, which rock composition(s) will you see from its base to
its summit? a. Mafic at the bottom, intermediate in the middle, and felsic at the top b. Felsic at
the bottom, intermediate in the middle, and mafic at the top c. Intermediate from bottom to top d.
Felsic from bottom to top
Solution
option (a) is correct
A . Mafic at the bottom ,
intermediate in the middle ,
and felsic at the top ..
What image of society is held by the three major theoretical approac.pdf
What image of society is held by the three major theoretical approaches? What core questions
does each approach ask? Which of the three approaches strikes you as the most useful? Why?
Solution
The three major theoretical approaches:-
1) Symbolic interactionism
2) Conflict Theory
3) Functionalism
Explanation:-
1) Symbolic interactionism:- emphasises on the language and symbols that help us to give
meaning to the experiences in our life. It is basically slices of society. It is distinctive approach to
the study of human life and human conduct.
2) Conflict Theory:- This theory argues that individuals and groups i.e., social classes within
society have different amounts of material and non-material resources (such as the wealthy vs.
the poor) and that the more powerful groups use their dominance and power in order to exploit
groups with less power.
3) Functionalism:-This theory focuses how everyone take a part in society, how is the behaviour
of the people in the society. It shows the social ethics of an individual. It distinguishes between
good and bad..
What are the major evolutionary trends that developedamong major v.pdf
What are the major evolutionary trends that developed
among major vertebrate groups, specifically those that
allowed for the transition from aquatic to terrestrial life?
Solution
Major evolutionary trends, specifically those that allowed for the transition from aquatic to
terrestrial life. These trends are developed respiration, protective and insulating body covering.
From transition from aquatic to terrestrial life changing to efficient reproductive methods like
placenta for some animals and also morphology of organism evolved such that for land they
would have paired , muscular appendages used for crawling and walking.
True or falsea. List index values must be integers.b. Lists can.pdf
True or false?
a. List index values must be integers.
b. Lists can change their size, getting larger or smaller.
c. A function cannot return a list.
d. All elements of a list are of the same type.
e. Lists cannot contain strings as elements.
f. A function cannot change the length of a list argument.
Solution
Below are True statements
a. List index values must be integers.
b. Lists can change their size, getting larger or smaller.
f. A function cannot change the length of a list argument.
Below are False statements
A function cannot return a list.
All elements of a list are of the same type.
e. Lists cannot contain strings as elements..
True false correction questions and fill in the blank. Same rules app.pdf
True false correction questions and fill in the blank. Same rules apply as to question 1 on the
true false part. Correct errors in the first part of the sentence (prior to the semicolon) Klebsiella
pneumonia causes COPD and urinary tract infections; the gram + organism also causes urinary
tract infections and can be part of the flora in this part of the body (give part of body).
Solution
true
Klebsiella pneumoniae causes chronic obstructive pulmonary disorder and urinary tract
infections.
the gram positive organism Lactobacillus delbrueckii also causes urinary tract infections and can
be part of flora in Vvagina of female body.
Show that all the singular points of the following functions are pol.pdf
Show that all the singular points of the following functions are poles. Determine the order of
each pole and the corresponding residue.
z + 1/z2 ? 2z
tanh z
z/cos z
Find the residue at z = 0 for the following functions:
csc2 z z-3
csc(z22)
zcos(1/z)
Solution
the pole i found was z = ?/2 + 2n? and -?/2 +2n? i attempted to find the residues using limits.
limz??/2 + 2n? (z-(?/2 + 2n?)) z/cosz doing some annoying algebra and just evaluating at ?/2 +
2n? i get 0/0. so i can use L\'hopitals rule, right? doing so i get my limit to equal ?/2 + 6n?.
The article at those photos groups 20 amino acids into 6 groups acco.pdf
The article at those photos groups 20 amino acids into 6 groups according to 8 amino acid
properties (size, partial volume, hydrophobicity, etc.). Perform the PCA analysis and show the
2D plot of 20 AAs according to the largest two principal components. the original strands and
one newly synthesized n RNA replication merely lead to mistakes in a sin strand that is
complementary t it. Clearly, both gle short-lived mRNA. Hence accurate DNA replica strands of
DNA contain the ful nformation n tion is very important. sary to recreate the other strand. The
key processes called this section \"closing the loop\" because of DNA replication occur at a
replication fork (Fi in the order that we presented things here, DNA 2.7(d)). At this point, the
two old strands are separ- replication is the last link in the cycle of mechan ated from one another
and the new strands are syn sms for synthesis of the major biological macro- thesized. The main
enzyme that does this job is DNA molecules. There is, however, a more fund lymerase III. This
enzyme catalyzesthe addition of sense in which this whole process is a loop. Clearly nucleotides
to the 3\'ends of the growing strands (at proteins cannot be synthesized without DN the heads of
the arrows in Fig 2.7(d)). The new cause proteins do not store genetic information DNA can
store this information, but it cannot carry strand is therefore synthesized in the 5\' to 3\' direc-
tion (as with mRNA synthesis during transcription) out the catalytic roles necessary for
metabolism in On one strand, called the leading strand, synthesis and it cann ot replicate itself
without the aid a cel is possible in a continuous unbroken fashion. How of proteins. There is thus
a chicken and egg situ ever, on the lagging strand on the opposite side, con ation: \"Which came
first, DNA or proteins Many tinuous synthesis is not possible and it is necessary peopl e now
believe that RNA preceded both DNA to initiate synthesis independently many times The and
proteins, and that there was a period in the new strand is therefore formed in pieces, which are
Earth\'s history when RNA played both the genetic known as Okazaki fragments and catalytic
roles. This is a tempting hypothesis DNA polymerase III is able to carry out theaddition because
several types of catalytic RNA are known of new nucleotides to a strand but it cannot initiate a
(both naturally occurring and artificially synthes- new strand. This is in contrast to RNA
polymerase ized sequences and because many viruses use which is able to perform both initiation
and addition RNA as their genetic materialtoday As with all con DNA polymerase therefore
needs a short sequence jectures related to the origin of life and very early called a primer, from
which to begin. Prime evolution, however, it is difficult to prove that an short sequences of RNA
ndicated by dotted lines in RNA world once existed. Fig. 2.7(d) that are synthesized by a form of
RNA polymerase called primase The processes of DNA s.
Studying Viruses can be important because they are often disease .pdf
Studying Viruses can be important because they are often disease causing pathogens. Howeve,
because of how easy they are to control and mainpulate in the laboratory, viruses can be good
tools to study basic, fundamental principles of cell and molecular biology. Concisely explain one
example of how viruses were used to make a critical molecular genetics discovery (briefly, how
was the experiment done, and what was concluded?)
Studying Viruses can be important because they are often disease causing pathogens. Howeve,
because of how easy they are to control and mainpulate in the laboratory, viruses can be good
tools to study basic, fundamental principles of cell and molecular biology. Concisely explain one
example of how viruses were used to make a critical molecular genetics discovery (briefly, how
was the experiment done, and what was concluded?)
Solution
Viruses are intracellular parasites that are effective after infecting particular host cells. Viral
infection starts offevolved when proteins on the surface of a virion bind to particular receptor
proteins at the host cells. The specificity of this interaction determines the host range of an
endemic.
Aside from being the causative agents of many sicknesses, viruses are vital gear in cellular
biology research, mainly in research on macromolecular synthesis.
Individual viral particles (virions) usually incorporate both an RNA or a DNA genome,
surrounded via multiple copies of 1 or a small variety of coat proteins, forming the nucleocapsid.
The nucleocapsid of many animal viruses is surrounded with the aid of a phospholipid bilayer, or
envelope.
During lytic replication, host-cell ribosomes and enzymes are used to explicit viral proteins,
which then viral genome and package it into viral coats. The multiple progeny virions produced
within a single infected cellular eventually are launched, following cellular lysis or gradual
disintegration of the cell . Progeny nucleocapsids of enveloped viruses are launched by using
budding of the host-cell membrane in which viral membrane proteins have been deposited.
Some bacterial viruses (bacteriophages) might also go through lysogeny following infection of
host cells. In this situation, the viral genome is incorporated into host-cell chromosomes, forming
a prophage this is replicated along with the host genome. When suitably activated, a prophage
enters the lytic cycle.
All retroviruses and some different animal viruses can combine their genomes into host-cell
chromosomes . In a few instances, this leads to abnormal cell replication and the eventual
improvement of cancers.
Recombinant viruses can be used as vectors to transduce genes into cells. In this method, viral
genes required for the lytic cycle are changed with the aid of other genes. The use of viral
vectors for gene remedy continues to be in its infancy, but has super ability for treatment of
diverse diseases..
State true or false with reason (all from Complex Analysis)-(a) P.pdf
State true or false with reason (all from Complex Analysis):-
(a) Power series has an antiderivative everywhere inside its disc of convergence
(c) There does not exist any analytic function f defined on C\\ {0} such that f \' (z) = 1/z
(d) If a mobius transformation T carries z1 and z2 into a same number w1 then either z1 = z2 or
else T is a constant map
Solution
(a) Power series has an antiderivative everywhere inside its disc of convergence. (TRUE).
Question 16What is a common tactic hackers use to allow themselves.pdf
Question 16
What is a common tactic hackers use to allow themselves re-access to a system?
Keystroke capture
Trap door
Backdoor
Rootkit
2.5 points
Question 17
What types of viruses propagate through the Internet with no user intervention?
Virus hoaxes
Stealth viruses
Worms
Trojan Horses
2.5 points
Question 18
A Web merchant wants to ensure that people cannot deny that they entered into a transaction, so
they starting using digital signatures and encryption as part of their transaction processing. This
is an example of what?
Integrity
Confidentiality
Authenticity
Non-repudiation
2.5 points
Question 19
What is the main solution used to prevent unauthorized access to systems, data, and applications?
Passwords
User profiles
Biometrics
Firewalls
2.5 points
Question 20
Which of the following refers to security solutions that protect access to systems, applications, or
data by ensuring users or systems are authenticated and allowed to have such access?
Physical access controls
User Profiles
Logical access controls
System access controlsA.
Keystroke captureB.
Trap doorC.
BackdoorD.
Rootkit
Solution
16. (D) Rootkit
17. (C) Worms
18. (D) Non-repudiation
19. (D) Firewalls
20. (D) System access controls.
Qualified plan documents can be written to accommodate plan forfeitu.pdf
Qualified plan documents can be written to accommodate plan forfeitures in different ways.
Which of the following statements is correct?
A. Stock bonus plans always allocate plan forfeitures to participants’ accounts since it is a stock
plan.
B. Safe harbor 401(k) plans never allocate plan forfeitures of matching contributions to
participants’ accounts.
C. Defined benefit plans (but not cash balance plans) have two options to deal with forfeitures.
D. Employers can use the forfeited funds to pay other expenses of the business outside the
qualified plan.
Solution.
Proteins fold. How WhySolutionANSProtein folding Protein.pdf
Proteins fold. How? Why?
Solution
ANS:
Protein folding: Proteins are the molecules in the body which carries out many important
functions, such as enzymes and antibodies.
Before proteins carry out their function, they must first assemble themselves or fold.
PROTEIN FOLDING MODELS:
1. Frame work model:
In this the protein local elements which are stretches of amino acids liked by one another by
covalently with peptide bonds. The region of the primary amino acids sequence in peptide bonds
make different domains.in this domains some of regions are made with Alfa - helix and some are
beta sheet domains.
These primary elements form secondary structure independently of tertiary structure (kim and
baldwin).these elements can diffuse until they collided, successfully adhering and coalescing to
give the tertiary structure (diffus-collision model), (karplus and weaven).
2. The classic nucleation model:
It postulates that some neighboring residues in the sequence would form native secondary
structure that would act as a nucleus form which the native structure would propagate in a step
wise manner. Thus the tertiary structure would form as necessary consequence of the secondary
structure (wetlaufer).
3. Hydrophobic collapse model:
It is hypothesized that a protein would collapse rapidly around its hydrophobic side chain and
then arrange from restricted conformational space occupied by the intermediate. Here the
secondary structure would be directed by native like tertiary structure (ptitsyn & kwajima).
Leviathal’s paradox:
It is formulated by Cyrus Leviathal. It explains, there are vast too many different possible
conformations for a protein to fold by a random search.
Structural organization of protein:
1. Primary Structure: the sequence of amino acids along with the di sulphide bonds.
2. SECONDARY STRUCTURE:
Alfa Helices and Beta- Pleated sheets are most frequently occur in secondary structure in
polypeptides. Both of these can case local folding of protein.
The Alfa Helices in polypeptide made up of L- Alfa amino acids are right handed, the Alfa -
Helices is stabilized by hydrogen bonds formed between oxygen and amide hydrogen’s of
peptide bonds that are a part of linear sequence.
Beta- Pleated sheets: 2 types
A. Parallel: in this the adjacent hydrogen bonding segments are connected by cross over
connection [Right handed].
B. Anti-Parallel: In this the adjacent hydrogen bonding segments are connected by HAIRPIN
turns or reverse turn [Left Handed].
3. Tertiary structure: It is arisen because of the folding of poly peptides upon itself.
4. Quaternary structure:
Some proteins are made up of 2 or more polypeptides chains, the interaction between different
poly peptide are subunits gives rise to the quaternary structure..
PLN 360560 Environmental Impact Assessment Homework No. 4 1. An appl.pdf
PLN 360/560 Environmental Impact Assessment Homework No. 4 1. An applicant has applied
to the U.S. Army Corps of Engineers (USACE) for a Clean Water Act (CWA) Section 404
Permit to construct a marina at the mouth of Chaotic Creek on Bagel Bay on Lake Promise, a
Water of the United States and Class \"B\" surface water in New York State. This site is located
in the upper reach of a large commercial harbor used for deep-draft navigation. The project
entails (1) construction of a breakwall; (2) dredging (and future maintenance dredging) of
navigation entrance and creek channels; (3) installation of 50 docks with boat slips; and (4)
construction of a boat launching ramp. During construction, it is estimated that a total of 60,000
cubic yards of material will be dredged and need to be managed, and about 2,500 cubic yards of
fill will need to be placed in Lake Promise. Management alternatives for the dredged material
include placement at an upland permitted municipal landfill, or discharge at a designated open-
water site in the harbor. The site is an abandoned \"hole\" and since discharge of the dredged
sediment is conceived to have benefits through reducing DDE bioaccumulation through capping
and restoration of shallow water aquatic habitat, it is referred to as a habitat restoration site
(HRS). The applicant had a consultant sample the sediments to be dredged for the marina as well
as a lake reference area and HRS, and had the samples analyzed by a laboratory. The following
is a summary of the test results that are to be submitted with the application: Contaminant
Concentration in sediment sample of Concentration in Concentration in reference bulk HRS bulk
marina sediment to be sediment sample sediment sample dredged Bulk Standard sediment
elutriate (mg/kg)(mg/L) (mg/kg) (mg/kg) 13 0.02 Barium Cyanide Iron Mercury Dieldrin p.p\'-
DDE Chlorobenzene 16 65 0.70 13,000 120 14 53 0.40 12,500 130 0.60 11,000 100 0.02 3x10-7
8x10 1.3x10 0.32 0.008 0.0034 0.12 1.3 38,000 Total organic 21,000 10,300 carbon (TOC) a.
What type of CWA regulatory approvals (i.e., permits, etc.) would be required for this project?
(2 pts.)
Solution
Approvals required for this project -
Storm Water permit - Clean Water Act (CWA) permit coverage for discharge of storm water
runoff from the construction site which are issued throught the US Environmental protection
agency\'s NPDES (national pollutant discharge elimination system) or the state NPDES authority
Permit under section 404 of the CWA issued by US Army Crops for Engineers
Resource Conservation and Recovery Act (RCRA) permit for generators of hazardous wastes
(The RCRA regulations list quantities of hazardous wastes that determine whether a generator is
large or small.), Storage of hazardous wastes (The RCRA regulations specify the time hazardous
wastes can be stored at a site. If the storage time is exceeded, a RCRA permit is required) and
Transport, treatment, and disposal of hazardous wastes (a transporter should be a .
one of the main design goals for the java programming language is se.pdf
one of the main design goals for the java programming language is security. Explain, in detail
how java achieves this goal?
Solution
1.Security became an essential piece of a java language from the starting day itself.Additionally
it is an progressing design aim of the java community .An intrinsic security is observed ,hence
the java language is in much now-a-days which is the major cause.
2.Every action is possible through Java virtual machine(JVM) .it is mainly to establish the
attributes like trust,integrity,confidentiality..etc.
3.Java offers large compile perusal ,leads with secondary stage of examining .language attributes
motivates the programmers closely to good programming recurents. especially the memory
management representation is uncomplicated.it removes the total classes of programing mistakes
which causes a continous trouble to the C and C++ programmers.moreover the objects are
fabricated with a new operator.precised programmer defined pointer data types, pointer
arthematic and self-activated garbage collector is absent.
4.java language is manipulated in distributed domains that tells that the security is most
important including the security attributes which is intended in the java language and run time
system.
5.In case of network habitats ,the applications in java language are secured from interruption
with an unofficial code venturing to acheive alrady the scenes and occupy file systems..
Meyer and colleagues performed three rounds of X-ray crystallography.pdf
Meyer and colleagues performed three rounds of X-ray crystallography, each with a different
elastase crystal. For each experiment, they positioned the elastase crystal at a different angle.
Why might researchers have rotated elastase crystals to position them at different angles?
Select one:
a. Similar to when you get X-rayed in different positions, X-rays would produce different images
depending on the angle and position of the elastase crystal. Researchers likely rotated elastase
crystals to different angles to understand the conformation of this enzyme from all sides.
b. There was no reason for the researchers to rotate the elastase crystals. It would have been
sufficient to carry out one X-ray crystallography experiment on an elastase crystal.
c. Rotating the elastase crystals to different angles likely affected the rate of the enzymatic
reaction; at a different angle, the hexapeptide substrate could remain bound to elastase for a
longer period of time. There was thus more time to get detailed X-ray images.
Solution
a.Similar to when you get X-rayed in different positions,X-rays would produce different images
depending on the angle and the position of the elactase crystal.Researchers likely rotated elastase
crystals to different angles and position of the elactase crystal.Researchers likely rotated elactase
crystals to different angles to understand to confirmation of this enzyme from all sides..
Numerous diseases are caused by problems in DNA replication or repai.pdf
Numerous diseases are caused by problems in DNA replication or repair. For example, Bloom
Syndrome is caused by a mutation in one of the helicase enzymes involved in DNA replication.
Select one of the enzymes involved in DNA replication or repair and find a disease related to a
mutation in this enzyme (a few of these diseases are mentioned in your textbook). Then, in at
least 3 pages, address the following topics: 1. What is the specific mutation? 2. What are the
symptoms of the disease? 3. How was the relationship between the mutation and the disease
discovered? 4. How does this mutation impact DNA replication and repair and how is this
different from normal DNA replication? 5. How does this mutation impact transcription and
translation and how is this different from normal transcription and translation? 6. How does this
problem in DNA replication lead to one symptom of the disease? 7. Give two examples of how
genetic engineering technology could be used to learn more about, detect earlier, or treat this
disease? Be sure to include a description of what genetic engineering is. You should write this
Assignment as an integrated paper that addresses all the questions. This Assignment should not
be done in question and answer format. Be sure to use appropriate headings for each section of
the paper.
Solution
Disease caused :Progressive external ophthalmoplegia (PEO) is caused by mutations in
mitochondrial DNA.
1.Specific mutation:Human mitochondrial DNA replication takes place by two-subunitsof DNA
polymerase gamma (polgamma).Mutations in the polymerase and exonuclease domains of the
human polgamma is the cause behind the occurence of PEO.
2.Symptoms of the disease:
3.Discovery of the relationship between DNA replication and repair: The diseases was
discovered the symptoms that the patients experience followed by a genetic test on muscle
biopsy sample.This led to the conclusion that there occured several deletions inmitochondrial
DNA.Research further revealed that there occured mutations in polgamma on chrmosome
15q25.In particular, three autosomal dominant mutations were revealed that altered the amino
acids located within polymerase motif B of polgamma.
4.Impact of mutation on DNA replication: The mutation resulted in dysfunctioning of the
polgamma and thus some regions of the mitochondrial DNA are deleted in comparision to
normal replication where the entire mitochondrial DNA is replicated fully.
5.Impact on transcription and translation: Mitochondrial DNA encodes 13 protein subunits of
multimeric oxidative phosphorylation, 2 ribosomal RNAs and all the translated RNAs that are
required for the translation of its own messenger RNA.In presence of mutations, some regions of
mitochindrial DNA is not replicated and this leads to impaired protein synthesis.
6.Problem in DNA replication lead to one symptom of the disease: Deletions in segments of
mitochondrial DNA leads to defective function of oxidative phosphorylation due to impaired
protein synth.
In the Mexican swordtail fish, the wild-type color is olive-green. Ot.pdf
In the Mexican swordtail fish, the wild-type color is olive-green. Other color variants include
golden and Take caution with ratio there is some reduced viability in this system. Matings
involving fish of these three phenotypes are shown below: Explain the inheritance of these color
phenotypes in the fish
Solution
At any rate, the mode of inheritance of albinism in the goldfish has not hitherto been known.
Using mates of different varieties of the goldfish, scientists arrived at the conclusion that
albinism of the goldfish is not a simple Mendelian character but it is represented by the double
homozygote of two non-linked autosomal recessive genes since because the F2 generation shows
15:1 ratio for non-albino and albino types like we see in the case of duplicate genes.
example: If the animal was A/a, because a is recessive and A is dominant, the animal\'s
phenotype wouldn\'t be albino, but it would still carry the gene for albinism (since it has an
\"a\"). Since it carries both \"A\" and \"a\", it is known as \"heterozygous\". If the animal had the
A/A combination, its phenotype wouldn\'t be albino, and it wouldn\'t carry the gene for albinism
(both genes being the same, it is called \"homozygous\" for \"A\"). If it were homozygous for
\"a\" (i.e. a/a), the animal\'s phenotype would be albino. Since \"a\" is recessive, both alleles need
to be \"a\" in order for albinism to be expressed in the phenotype.
The genes for the pigment cells are inherited independently of one another, and there is no
known linkage to any other genes. So, each pigment type is coded by two different genes, one on
each of a pair of chromosomes. The phenotype of wild type animals is dark, non-melanoid, non-
albino and non-axanthic.
The golden colour (D/D a/a or D/d a/a). It has normal migration of pigment cells, but lacks
melanophores, hence the yellow/gold appearance..
If we maintain a strict biological definition of “species”, H. sapie.pdf
If we maintain a strict biological definition of “species”, H. sapiens and H. neanderthalensis were
NOT separate species as they successfully interbred. What impact might this information have
on assigning a date to the origin of humans? Are there any other implications?
Solution
As we know the common male ancester is usually much younger than the common female
ancester.Moreover the Mitochondrial DNA which is present in the Neanderthals and it is absent
in humans
Suppose we have related populations A, B and C and we can have that A can interbreed with B,
and also the B can interbreed with C,.At the same time ,A can not interbreed with C. Noe a
ccording to the strict BSC, the A and B are in the same species and at the same time B and C are
in the same species, but A and C are not.These are the implications
here.
It is impossible to know the date of origins of humans due to difference between sapiens and
neanderthals.we can fix one date on which an old species was found and can assume it to be the
date of origin..
If taxpayers Monica, Traci , and Phillip are responsible persons in .pdf
If taxpayers Monica, Traci , and Phillip are responsible persons in the failure to withhold and pay
over employment taxes but only Monica and Traci acted willfully which of the following
statements is correct? A the IRs may assess the trust fund recovery penalty against Monica ,
Traci and Phillip, but can only collect it from Monica and Traci b. The IRS can only asses and
collect a trust fund recovery penalty from Phillip c. The IRS cannot assess a trust find recovery
penalty against Phillip d. the IRS may asses a trust fund recovery penalty against Monica , Traci,
and Phillip
Solution
The tax IRs may assess a trust recovery penalty against monica, traci, and phillip, but can only
collect it from monica and trici.
The employer need to withhold taxes from its employees wages and pay them to IRS. Withheld
taxes are been said as trust fund taxes. When these taxes are not paid by the employer
intenstionally or not to get paid the taxes IRs checks for the responsible person and willful
person and upon all IRs may assess for thr trust fund recovery. IRs may ask for to pay the
penality by any of the responsible persons..
human biologyscientific method projectStep 1 Observe and Genera.pdf
human biology
scientific method project
Step 1: Observe and Generalize ----common cold example hand washing name two more
Observe different characteristics related to your topic of question, and then generalize about it.
Take into consideration what is already known on the subject, varying opinions, past mistakes,
and new aspects that you may have not foreseen. Summarize:
Observation 1:
Observation 2:
Observation 3:
Solution
Answer:
Observation 1: Common cold is caused by bacterial or viral infection and transmitted from one
person to other through droplets from infected person. The person infected with common cold
show protect his nose or mouth with steril mask, which covers mouth and nose. If nose or mouth
is not covered by mask, droplets can transmit common cold to many people in surrounding
environment.
Observation 2: Common cold is widely transmitted very easily, if a person infected with
common cold and he sneezes in public places which is having high population, several thousands
of droplets can be transported through air at onetime, when person sneezes, so avoid moving in
public places when common cold is rampant in specific area.
Observation 3: Doctors who examine the persons suffering with common cold should be careful,
the physician should use antibiotics or vaccines to prevent common cold infection..
I would appreciate help with these questions. Thank you.Question 1.pdf
I would appreciate help with these questions. Thank you.
Question 1
How long is the typical default keepalive timer for TCP?
10 milliseconds
1 second
5 minutes
2 hours
Question 2
What is the primary reason that large packets may not be more efficient in routing IP networks?
The additional processing time associated with the cut-through nature of the IP router
The additional overhead associated with a larger bandwidth connection
The additional latency due to the store-and-forward nature of the IP router
The TCP connection overhead required from hop-to-hop
Question 3
What additional parameter is typically used to protect against wrapped sequence numbers on
high-bandwidth connections?
keepalive
TTL
timestamp
DF
Question 4
What does OID stand for?
Object IDentifier
Object Instance Data
Organized Identifying Data
Ordered Information Director
Question 5
What is the typical port for an SNMP agent to use for managers to connect to?
80
161
22
1812
Question 6
Which of the following are notifications generated by the snmp agent on a router, rather than
information requested by an snmp manager?
get
get-bulk
get-next
trap1.
10 milliseconds2.
1 second3.
5 minutes4.
2 hours
Solution
Answer :
Question 1
How long is the typical default keepalive timer for TCP?
Answer :
4. 2 hours
2 hours is the typical default keepAlive time for TCP
Question 3
What additional parameter is typically used to protect against wrapped sequence numbers on
high-bandwidth connections?
Answer :
3.timestamp
timestamp parameter is typically used to protect against wrapped sequence numbers on high-
bandwidth connections
Question 4
What does OID stand for?
Answer :
1.Object IDentifier
OID stand for Object IDentifier
Question 5
What is the typical port for an SNMP agent to use for managers to connect to?
Answer :
2.161
161 is the typical port for an SNMP agent to use for managers to connect to.
Question 6
Which of the following are notifications generated by the snmp agent on a router, rather than
information requested by an snmp manager?
Answer :
4.trap
Traps are notifications generated by the snmp agent on a router, rather than information
requested by an snmp manager..
For settlements, what are the major components of a soil settlment W.pdf
For settlements, what are the major components of a soil settlment? What is the difference
between between settlment in sands and clays, please describe the details. 2.
Solution
settlement have three components: 1. Immediate settlement, 2. Primary settlement and
3.secondary settlement.
immediate settlement takes place immediately after the application of load.
Primary settlement is the long term process which occurs after the application of load.
secondary settlement depends on time.
In sands, immediate settlement takes place and in clays primary settlement occurs..
The approach at University of Liverpool.pptx
How libraries can support authors with open access requirements for UKRI funded books
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What are the major evolutionary trends that developedamong major v.pdf
What are the major evolutionary trends that developed
among major vertebrate groups, specifically those that
allowed for the transition from aquatic to terrestrial life?
Solution
Major evolutionary trends, specifically those that allowed for the transition from aquatic to
terrestrial life. These trends are developed respiration, protective and insulating body covering.
From transition from aquatic to terrestrial life changing to efficient reproductive methods like
placenta for some animals and also morphology of organism evolved such that for land they
would have paired , muscular appendages used for crawling and walking.
True or falsea. List index values must be integers.b. Lists can.pdf
True or false?
a. List index values must be integers.
b. Lists can change their size, getting larger or smaller.
c. A function cannot return a list.
d. All elements of a list are of the same type.
e. Lists cannot contain strings as elements.
f. A function cannot change the length of a list argument.
Solution
Below are True statements
a. List index values must be integers.
b. Lists can change their size, getting larger or smaller.
f. A function cannot change the length of a list argument.
Below are False statements
A function cannot return a list.
All elements of a list are of the same type.
e. Lists cannot contain strings as elements..
True false correction questions and fill in the blank. Same rules app.pdf
True false correction questions and fill in the blank. Same rules apply as to question 1 on the
true false part. Correct errors in the first part of the sentence (prior to the semicolon) Klebsiella
pneumonia causes COPD and urinary tract infections; the gram + organism also causes urinary
tract infections and can be part of the flora in this part of the body (give part of body).
Solution
true
Klebsiella pneumoniae causes chronic obstructive pulmonary disorder and urinary tract
infections.
the gram positive organism Lactobacillus delbrueckii also causes urinary tract infections and can
be part of flora in Vvagina of female body.
Show that all the singular points of the following functions are pol.pdf
Show that all the singular points of the following functions are poles. Determine the order of
each pole and the corresponding residue.
z + 1/z2 ? 2z
tanh z
z/cos z
Find the residue at z = 0 for the following functions:
csc2 z z-3
csc(z22)
zcos(1/z)
Solution
the pole i found was z = ?/2 + 2n? and -?/2 +2n? i attempted to find the residues using limits.
limz??/2 + 2n? (z-(?/2 + 2n?)) z/cosz doing some annoying algebra and just evaluating at ?/2 +
2n? i get 0/0. so i can use L\'hopitals rule, right? doing so i get my limit to equal ?/2 + 6n?.
The article at those photos groups 20 amino acids into 6 groups acco.pdf
The article at those photos groups 20 amino acids into 6 groups according to 8 amino acid
properties (size, partial volume, hydrophobicity, etc.). Perform the PCA analysis and show the
2D plot of 20 AAs according to the largest two principal components. the original strands and
one newly synthesized n RNA replication merely lead to mistakes in a sin strand that is
complementary t it. Clearly, both gle short-lived mRNA. Hence accurate DNA replica strands of
DNA contain the ful nformation n tion is very important. sary to recreate the other strand. The
key processes called this section \"closing the loop\" because of DNA replication occur at a
replication fork (Fi in the order that we presented things here, DNA 2.7(d)). At this point, the
two old strands are separ- replication is the last link in the cycle of mechan ated from one another
and the new strands are syn sms for synthesis of the major biological macro- thesized. The main
enzyme that does this job is DNA molecules. There is, however, a more fund lymerase III. This
enzyme catalyzesthe addition of sense in which this whole process is a loop. Clearly nucleotides
to the 3\'ends of the growing strands (at proteins cannot be synthesized without DN the heads of
the arrows in Fig 2.7(d)). The new cause proteins do not store genetic information DNA can
store this information, but it cannot carry strand is therefore synthesized in the 5\' to 3\' direc-
tion (as with mRNA synthesis during transcription) out the catalytic roles necessary for
metabolism in On one strand, called the leading strand, synthesis and it cann ot replicate itself
without the aid a cel is possible in a continuous unbroken fashion. How of proteins. There is thus
a chicken and egg situ ever, on the lagging strand on the opposite side, con ation: \"Which came
first, DNA or proteins Many tinuous synthesis is not possible and it is necessary peopl e now
believe that RNA preceded both DNA to initiate synthesis independently many times The and
proteins, and that there was a period in the new strand is therefore formed in pieces, which are
Earth\'s history when RNA played both the genetic known as Okazaki fragments and catalytic
roles. This is a tempting hypothesis DNA polymerase III is able to carry out theaddition because
several types of catalytic RNA are known of new nucleotides to a strand but it cannot initiate a
(both naturally occurring and artificially synthes- new strand. This is in contrast to RNA
polymerase ized sequences and because many viruses use which is able to perform both initiation
and addition RNA as their genetic materialtoday As with all con DNA polymerase therefore
needs a short sequence jectures related to the origin of life and very early called a primer, from
which to begin. Prime evolution, however, it is difficult to prove that an short sequences of RNA
ndicated by dotted lines in RNA world once existed. Fig. 2.7(d) that are synthesized by a form of
RNA polymerase called primase The processes of DNA s.
Studying Viruses can be important because they are often disease .pdf
Studying Viruses can be important because they are often disease causing pathogens. Howeve,
because of how easy they are to control and mainpulate in the laboratory, viruses can be good
tools to study basic, fundamental principles of cell and molecular biology. Concisely explain one
example of how viruses were used to make a critical molecular genetics discovery (briefly, how
was the experiment done, and what was concluded?)
Studying Viruses can be important because they are often disease causing pathogens. Howeve,
because of how easy they are to control and mainpulate in the laboratory, viruses can be good
tools to study basic, fundamental principles of cell and molecular biology. Concisely explain one
example of how viruses were used to make a critical molecular genetics discovery (briefly, how
was the experiment done, and what was concluded?)
Solution
Viruses are intracellular parasites that are effective after infecting particular host cells. Viral
infection starts offevolved when proteins on the surface of a virion bind to particular receptor
proteins at the host cells. The specificity of this interaction determines the host range of an
endemic.
Aside from being the causative agents of many sicknesses, viruses are vital gear in cellular
biology research, mainly in research on macromolecular synthesis.
Individual viral particles (virions) usually incorporate both an RNA or a DNA genome,
surrounded via multiple copies of 1 or a small variety of coat proteins, forming the nucleocapsid.
The nucleocapsid of many animal viruses is surrounded with the aid of a phospholipid bilayer, or
envelope.
During lytic replication, host-cell ribosomes and enzymes are used to explicit viral proteins,
which then viral genome and package it into viral coats. The multiple progeny virions produced
within a single infected cellular eventually are launched, following cellular lysis or gradual
disintegration of the cell . Progeny nucleocapsids of enveloped viruses are launched by using
budding of the host-cell membrane in which viral membrane proteins have been deposited.
Some bacterial viruses (bacteriophages) might also go through lysogeny following infection of
host cells. In this situation, the viral genome is incorporated into host-cell chromosomes, forming
a prophage this is replicated along with the host genome. When suitably activated, a prophage
enters the lytic cycle.
All retroviruses and some different animal viruses can combine their genomes into host-cell
chromosomes . In a few instances, this leads to abnormal cell replication and the eventual
improvement of cancers.
Recombinant viruses can be used as vectors to transduce genes into cells. In this method, viral
genes required for the lytic cycle are changed with the aid of other genes. The use of viral
vectors for gene remedy continues to be in its infancy, but has super ability for treatment of
diverse diseases..
State true or false with reason (all from Complex Analysis)-(a) P.pdf
State true or false with reason (all from Complex Analysis):-
(a) Power series has an antiderivative everywhere inside its disc of convergence
(c) There does not exist any analytic function f defined on C\\ {0} such that f \' (z) = 1/z
(d) If a mobius transformation T carries z1 and z2 into a same number w1 then either z1 = z2 or
else T is a constant map
Solution
(a) Power series has an antiderivative everywhere inside its disc of convergence. (TRUE).
Question 16What is a common tactic hackers use to allow themselves.pdf
Question 16
What is a common tactic hackers use to allow themselves re-access to a system?
Keystroke capture
Trap door
Backdoor
Rootkit
2.5 points
Question 17
What types of viruses propagate through the Internet with no user intervention?
Virus hoaxes
Stealth viruses
Worms
Trojan Horses
2.5 points
Question 18
A Web merchant wants to ensure that people cannot deny that they entered into a transaction, so
they starting using digital signatures and encryption as part of their transaction processing. This
is an example of what?
Integrity
Confidentiality
Authenticity
Non-repudiation
2.5 points
Question 19
What is the main solution used to prevent unauthorized access to systems, data, and applications?
Passwords
User profiles
Biometrics
Firewalls
2.5 points
Question 20
Which of the following refers to security solutions that protect access to systems, applications, or
data by ensuring users or systems are authenticated and allowed to have such access?
Physical access controls
User Profiles
Logical access controls
System access controlsA.
Keystroke captureB.
Trap doorC.
BackdoorD.
Rootkit
Solution
16. (D) Rootkit
17. (C) Worms
18. (D) Non-repudiation
19. (D) Firewalls
20. (D) System access controls.
Qualified plan documents can be written to accommodate plan forfeitu.pdf
Qualified plan documents can be written to accommodate plan forfeitures in different ways.
Which of the following statements is correct?
A. Stock bonus plans always allocate plan forfeitures to participants’ accounts since it is a stock
plan.
B. Safe harbor 401(k) plans never allocate plan forfeitures of matching contributions to
participants’ accounts.
C. Defined benefit plans (but not cash balance plans) have two options to deal with forfeitures.
D. Employers can use the forfeited funds to pay other expenses of the business outside the
qualified plan.
Solution.
Proteins fold. How WhySolutionANSProtein folding Protein.pdf
Proteins fold. How? Why?
Solution
ANS:
Protein folding: Proteins are the molecules in the body which carries out many important
functions, such as enzymes and antibodies.
Before proteins carry out their function, they must first assemble themselves or fold.
PROTEIN FOLDING MODELS:
1. Frame work model:
In this the protein local elements which are stretches of amino acids liked by one another by
covalently with peptide bonds. The region of the primary amino acids sequence in peptide bonds
make different domains.in this domains some of regions are made with Alfa - helix and some are
beta sheet domains.
These primary elements form secondary structure independently of tertiary structure (kim and
baldwin).these elements can diffuse until they collided, successfully adhering and coalescing to
give the tertiary structure (diffus-collision model), (karplus and weaven).
2. The classic nucleation model:
It postulates that some neighboring residues in the sequence would form native secondary
structure that would act as a nucleus form which the native structure would propagate in a step
wise manner. Thus the tertiary structure would form as necessary consequence of the secondary
structure (wetlaufer).
3. Hydrophobic collapse model:
It is hypothesized that a protein would collapse rapidly around its hydrophobic side chain and
then arrange from restricted conformational space occupied by the intermediate. Here the
secondary structure would be directed by native like tertiary structure (ptitsyn & kwajima).
Leviathal’s paradox:
It is formulated by Cyrus Leviathal. It explains, there are vast too many different possible
conformations for a protein to fold by a random search.
Structural organization of protein:
1. Primary Structure: the sequence of amino acids along with the di sulphide bonds.
2. SECONDARY STRUCTURE:
Alfa Helices and Beta- Pleated sheets are most frequently occur in secondary structure in
polypeptides. Both of these can case local folding of protein.
The Alfa Helices in polypeptide made up of L- Alfa amino acids are right handed, the Alfa -
Helices is stabilized by hydrogen bonds formed between oxygen and amide hydrogen’s of
peptide bonds that are a part of linear sequence.
Beta- Pleated sheets: 2 types
A. Parallel: in this the adjacent hydrogen bonding segments are connected by cross over
connection [Right handed].
B. Anti-Parallel: In this the adjacent hydrogen bonding segments are connected by HAIRPIN
turns or reverse turn [Left Handed].
3. Tertiary structure: It is arisen because of the folding of poly peptides upon itself.
4. Quaternary structure:
Some proteins are made up of 2 or more polypeptides chains, the interaction between different
poly peptide are subunits gives rise to the quaternary structure..
PLN 360560 Environmental Impact Assessment Homework No. 4 1. An appl.pdf
PLN 360/560 Environmental Impact Assessment Homework No. 4 1. An applicant has applied
to the U.S. Army Corps of Engineers (USACE) for a Clean Water Act (CWA) Section 404
Permit to construct a marina at the mouth of Chaotic Creek on Bagel Bay on Lake Promise, a
Water of the United States and Class \"B\" surface water in New York State. This site is located
in the upper reach of a large commercial harbor used for deep-draft navigation. The project
entails (1) construction of a breakwall; (2) dredging (and future maintenance dredging) of
navigation entrance and creek channels; (3) installation of 50 docks with boat slips; and (4)
construction of a boat launching ramp. During construction, it is estimated that a total of 60,000
cubic yards of material will be dredged and need to be managed, and about 2,500 cubic yards of
fill will need to be placed in Lake Promise. Management alternatives for the dredged material
include placement at an upland permitted municipal landfill, or discharge at a designated open-
water site in the harbor. The site is an abandoned \"hole\" and since discharge of the dredged
sediment is conceived to have benefits through reducing DDE bioaccumulation through capping
and restoration of shallow water aquatic habitat, it is referred to as a habitat restoration site
(HRS). The applicant had a consultant sample the sediments to be dredged for the marina as well
as a lake reference area and HRS, and had the samples analyzed by a laboratory. The following
is a summary of the test results that are to be submitted with the application: Contaminant
Concentration in sediment sample of Concentration in Concentration in reference bulk HRS bulk
marina sediment to be sediment sample sediment sample dredged Bulk Standard sediment
elutriate (mg/kg)(mg/L) (mg/kg) (mg/kg) 13 0.02 Barium Cyanide Iron Mercury Dieldrin p.p\'-
DDE Chlorobenzene 16 65 0.70 13,000 120 14 53 0.40 12,500 130 0.60 11,000 100 0.02 3x10-7
8x10 1.3x10 0.32 0.008 0.0034 0.12 1.3 38,000 Total organic 21,000 10,300 carbon (TOC) a.
What type of CWA regulatory approvals (i.e., permits, etc.) would be required for this project?
(2 pts.)
Solution
Approvals required for this project -
Storm Water permit - Clean Water Act (CWA) permit coverage for discharge of storm water
runoff from the construction site which are issued throught the US Environmental protection
agency\'s NPDES (national pollutant discharge elimination system) or the state NPDES authority
Permit under section 404 of the CWA issued by US Army Crops for Engineers
Resource Conservation and Recovery Act (RCRA) permit for generators of hazardous wastes
(The RCRA regulations list quantities of hazardous wastes that determine whether a generator is
large or small.), Storage of hazardous wastes (The RCRA regulations specify the time hazardous
wastes can be stored at a site. If the storage time is exceeded, a RCRA permit is required) and
Transport, treatment, and disposal of hazardous wastes (a transporter should be a .
one of the main design goals for the java programming language is se.pdf
one of the main design goals for the java programming language is security. Explain, in detail
how java achieves this goal?
Solution
1.Security became an essential piece of a java language from the starting day itself.Additionally
it is an progressing design aim of the java community .An intrinsic security is observed ,hence
the java language is in much now-a-days which is the major cause.
2.Every action is possible through Java virtual machine(JVM) .it is mainly to establish the
attributes like trust,integrity,confidentiality..etc.
3.Java offers large compile perusal ,leads with secondary stage of examining .language attributes
motivates the programmers closely to good programming recurents. especially the memory
management representation is uncomplicated.it removes the total classes of programing mistakes
which causes a continous trouble to the C and C++ programmers.moreover the objects are
fabricated with a new operator.precised programmer defined pointer data types, pointer
arthematic and self-activated garbage collector is absent.
4.java language is manipulated in distributed domains that tells that the security is most
important including the security attributes which is intended in the java language and run time
system.
5.In case of network habitats ,the applications in java language are secured from interruption
with an unofficial code venturing to acheive alrady the scenes and occupy file systems..
Meyer and colleagues performed three rounds of X-ray crystallography.pdf
Meyer and colleagues performed three rounds of X-ray crystallography, each with a different
elastase crystal. For each experiment, they positioned the elastase crystal at a different angle.
Why might researchers have rotated elastase crystals to position them at different angles?
Select one:
a. Similar to when you get X-rayed in different positions, X-rays would produce different images
depending on the angle and position of the elastase crystal. Researchers likely rotated elastase
crystals to different angles to understand the conformation of this enzyme from all sides.
b. There was no reason for the researchers to rotate the elastase crystals. It would have been
sufficient to carry out one X-ray crystallography experiment on an elastase crystal.
c. Rotating the elastase crystals to different angles likely affected the rate of the enzymatic
reaction; at a different angle, the hexapeptide substrate could remain bound to elastase for a
longer period of time. There was thus more time to get detailed X-ray images.
Solution
a.Similar to when you get X-rayed in different positions,X-rays would produce different images
depending on the angle and the position of the elactase crystal.Researchers likely rotated elastase
crystals to different angles and position of the elactase crystal.Researchers likely rotated elactase
crystals to different angles to understand to confirmation of this enzyme from all sides..
Numerous diseases are caused by problems in DNA replication or repai.pdf
Numerous diseases are caused by problems in DNA replication or repair. For example, Bloom
Syndrome is caused by a mutation in one of the helicase enzymes involved in DNA replication.
Select one of the enzymes involved in DNA replication or repair and find a disease related to a
mutation in this enzyme (a few of these diseases are mentioned in your textbook). Then, in at
least 3 pages, address the following topics: 1. What is the specific mutation? 2. What are the
symptoms of the disease? 3. How was the relationship between the mutation and the disease
discovered? 4. How does this mutation impact DNA replication and repair and how is this
different from normal DNA replication? 5. How does this mutation impact transcription and
translation and how is this different from normal transcription and translation? 6. How does this
problem in DNA replication lead to one symptom of the disease? 7. Give two examples of how
genetic engineering technology could be used to learn more about, detect earlier, or treat this
disease? Be sure to include a description of what genetic engineering is. You should write this
Assignment as an integrated paper that addresses all the questions. This Assignment should not
be done in question and answer format. Be sure to use appropriate headings for each section of
the paper.
Solution
Disease caused :Progressive external ophthalmoplegia (PEO) is caused by mutations in
mitochondrial DNA.
1.Specific mutation:Human mitochondrial DNA replication takes place by two-subunitsof DNA
polymerase gamma (polgamma).Mutations in the polymerase and exonuclease domains of the
human polgamma is the cause behind the occurence of PEO.
2.Symptoms of the disease:
3.Discovery of the relationship between DNA replication and repair: The diseases was
discovered the symptoms that the patients experience followed by a genetic test on muscle
biopsy sample.This led to the conclusion that there occured several deletions inmitochondrial
DNA.Research further revealed that there occured mutations in polgamma on chrmosome
15q25.In particular, three autosomal dominant mutations were revealed that altered the amino
acids located within polymerase motif B of polgamma.
4.Impact of mutation on DNA replication: The mutation resulted in dysfunctioning of the
polgamma and thus some regions of the mitochondrial DNA are deleted in comparision to
normal replication where the entire mitochondrial DNA is replicated fully.
5.Impact on transcription and translation: Mitochondrial DNA encodes 13 protein subunits of
multimeric oxidative phosphorylation, 2 ribosomal RNAs and all the translated RNAs that are
required for the translation of its own messenger RNA.In presence of mutations, some regions of
mitochindrial DNA is not replicated and this leads to impaired protein synthesis.
6.Problem in DNA replication lead to one symptom of the disease: Deletions in segments of
mitochondrial DNA leads to defective function of oxidative phosphorylation due to impaired
protein synth.
In the Mexican swordtail fish, the wild-type color is olive-green. Ot.pdf
In the Mexican swordtail fish, the wild-type color is olive-green. Other color variants include
golden and Take caution with ratio there is some reduced viability in this system. Matings
involving fish of these three phenotypes are shown below: Explain the inheritance of these color
phenotypes in the fish
Solution
At any rate, the mode of inheritance of albinism in the goldfish has not hitherto been known.
Using mates of different varieties of the goldfish, scientists arrived at the conclusion that
albinism of the goldfish is not a simple Mendelian character but it is represented by the double
homozygote of two non-linked autosomal recessive genes since because the F2 generation shows
15:1 ratio for non-albino and albino types like we see in the case of duplicate genes.
example: If the animal was A/a, because a is recessive and A is dominant, the animal\'s
phenotype wouldn\'t be albino, but it would still carry the gene for albinism (since it has an
\"a\"). Since it carries both \"A\" and \"a\", it is known as \"heterozygous\". If the animal had the
A/A combination, its phenotype wouldn\'t be albino, and it wouldn\'t carry the gene for albinism
(both genes being the same, it is called \"homozygous\" for \"A\"). If it were homozygous for
\"a\" (i.e. a/a), the animal\'s phenotype would be albino. Since \"a\" is recessive, both alleles need
to be \"a\" in order for albinism to be expressed in the phenotype.
The genes for the pigment cells are inherited independently of one another, and there is no
known linkage to any other genes. So, each pigment type is coded by two different genes, one on
each of a pair of chromosomes. The phenotype of wild type animals is dark, non-melanoid, non-
albino and non-axanthic.
The golden colour (D/D a/a or D/d a/a). It has normal migration of pigment cells, but lacks
melanophores, hence the yellow/gold appearance..
If we maintain a strict biological definition of “species”, H. sapie.pdf
If we maintain a strict biological definition of “species”, H. sapiens and H. neanderthalensis were
NOT separate species as they successfully interbred. What impact might this information have
on assigning a date to the origin of humans? Are there any other implications?
Solution
As we know the common male ancester is usually much younger than the common female
ancester.Moreover the Mitochondrial DNA which is present in the Neanderthals and it is absent
in humans
Suppose we have related populations A, B and C and we can have that A can interbreed with B,
and also the B can interbreed with C,.At the same time ,A can not interbreed with C. Noe a
ccording to the strict BSC, the A and B are in the same species and at the same time B and C are
in the same species, but A and C are not.These are the implications
here.
It is impossible to know the date of origins of humans due to difference between sapiens and
neanderthals.we can fix one date on which an old species was found and can assume it to be the
date of origin..
If taxpayers Monica, Traci , and Phillip are responsible persons in .pdf
If taxpayers Monica, Traci , and Phillip are responsible persons in the failure to withhold and pay
over employment taxes but only Monica and Traci acted willfully which of the following
statements is correct? A the IRs may assess the trust fund recovery penalty against Monica ,
Traci and Phillip, but can only collect it from Monica and Traci b. The IRS can only asses and
collect a trust fund recovery penalty from Phillip c. The IRS cannot assess a trust find recovery
penalty against Phillip d. the IRS may asses a trust fund recovery penalty against Monica , Traci,
and Phillip
Solution
The tax IRs may assess a trust recovery penalty against monica, traci, and phillip, but can only
collect it from monica and trici.
The employer need to withhold taxes from its employees wages and pay them to IRS. Withheld
taxes are been said as trust fund taxes. When these taxes are not paid by the employer
intenstionally or not to get paid the taxes IRs checks for the responsible person and willful
person and upon all IRs may assess for thr trust fund recovery. IRs may ask for to pay the
penality by any of the responsible persons..
human biologyscientific method projectStep 1 Observe and Genera.pdf
human biology
scientific method project
Step 1: Observe and Generalize ----common cold example hand washing name two more
Observe different characteristics related to your topic of question, and then generalize about it.
Take into consideration what is already known on the subject, varying opinions, past mistakes,
and new aspects that you may have not foreseen. Summarize:
Observation 1:
Observation 2:
Observation 3:
Solution
Answer:
Observation 1: Common cold is caused by bacterial or viral infection and transmitted from one
person to other through droplets from infected person. The person infected with common cold
show protect his nose or mouth with steril mask, which covers mouth and nose. If nose or mouth
is not covered by mask, droplets can transmit common cold to many people in surrounding
environment.
Observation 2: Common cold is widely transmitted very easily, if a person infected with
common cold and he sneezes in public places which is having high population, several thousands
of droplets can be transported through air at onetime, when person sneezes, so avoid moving in
public places when common cold is rampant in specific area.
Observation 3: Doctors who examine the persons suffering with common cold should be careful,
the physician should use antibiotics or vaccines to prevent common cold infection..
I would appreciate help with these questions. Thank you.Question 1.pdf
I would appreciate help with these questions. Thank you.
Question 1
How long is the typical default keepalive timer for TCP?
10 milliseconds
1 second
5 minutes
2 hours
Question 2
What is the primary reason that large packets may not be more efficient in routing IP networks?
The additional processing time associated with the cut-through nature of the IP router
The additional overhead associated with a larger bandwidth connection
The additional latency due to the store-and-forward nature of the IP router
The TCP connection overhead required from hop-to-hop
Question 3
What additional parameter is typically used to protect against wrapped sequence numbers on
high-bandwidth connections?
keepalive
TTL
timestamp
DF
Question 4
What does OID stand for?
Object IDentifier
Object Instance Data
Organized Identifying Data
Ordered Information Director
Question 5
What is the typical port for an SNMP agent to use for managers to connect to?
80
161
22
1812
Question 6
Which of the following are notifications generated by the snmp agent on a router, rather than
information requested by an snmp manager?
get
get-bulk
get-next
trap1.
10 milliseconds2.
1 second3.
5 minutes4.
2 hours
Solution
Answer :
Question 1
How long is the typical default keepalive timer for TCP?
Answer :
4. 2 hours
2 hours is the typical default keepAlive time for TCP
Question 3
What additional parameter is typically used to protect against wrapped sequence numbers on
high-bandwidth connections?
Answer :
3.timestamp
timestamp parameter is typically used to protect against wrapped sequence numbers on high-
bandwidth connections
Question 4
What does OID stand for?
Answer :
1.Object IDentifier
OID stand for Object IDentifier
Question 5
What is the typical port for an SNMP agent to use for managers to connect to?
Answer :
2.161
161 is the typical port for an SNMP agent to use for managers to connect to.
Question 6
Which of the following are notifications generated by the snmp agent on a router, rather than
information requested by an snmp manager?
Answer :
4.trap
Traps are notifications generated by the snmp agent on a router, rather than information
requested by an snmp manager..
For settlements, what are the major components of a soil settlment W.pdf
For settlements, what are the major components of a soil settlment? What is the difference
between between settlment in sands and clays, please describe the details. 2.
Solution
settlement have three components: 1. Immediate settlement, 2. Primary settlement and
3.secondary settlement.
immediate settlement takes place immediately after the application of load.
Primary settlement is the long term process which occurs after the application of load.
secondary settlement depends on time.
In sands, immediate settlement takes place and in clays primary settlement occurs..
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send edited codeMain 2public class Main2 { public static voi.pdf
- 1. send edited code Main 2 public class Main2 { public static void main(String[] args) { EZ.initialize(512, 512); EZ.addImage("metropolis.jpg", 256, 256); Hero myHero = new Hero(100,390); //HeroPlus mySecondHero = new HeroPlus(100,390); while(true){ myHero.processStates(); //mySecondHero.processStates(); EZ.refreshScreen(); } } } Hero Plus public class HeroPlus { EZImage heroStand; EZImage heroJump; EZImage heroLand; EZImage heroHover; EZImage heroPunch; // PUNCH int posx = 0; int posy = 0; int heroState = STAND; static final int STAND = 1; static final int JUMP = 2; static final int LAND = 3; static final int HOVER = 4; static final int PUNCH = 5; static final int JUMPHEIGHT = 100;
- 2. static final int PUNCHDISTANCE = 50; // PUNCH int punchCounter = 0; // PUNCH int jumpCounter = 0; HeroPlus(int x, int y) { posx = x; posy = y; heroStand = EZ.addImage("superstand.png", posx, posy); heroLand = EZ.addImage("superland.png", posx, posy); heroJump = EZ.addImage("superjump.png", posx, posy); heroHover = EZ.addImage("superhover.png", posx, posy); heroPunch = EZ.addImage("superpunch.png", posx, posy); // PUNCH hideHero(); heroStand.show(); } void hideHero() { heroStand.hide(); heroLand.hide(); heroJump.hide(); heroHover.hide(); heroPunch.hide(); // PUNCH } void positionHero(int x, int y) { heroStand.translateTo(x, y); heroJump.translateTo(x, y); heroLand.translateTo(x, y); heroHover.translateTo(x, y); heroPunch.translateTo(x, y); // PUNCH } void processStates() { switch (heroState) { case STAND: if (EZInteraction.wasKeyPressed('j')) { heroState = JUMP; jumpCounter = 0; hideHero(); heroJump.show();
- 3. } break; case JUMP: jumpCounter++; if (EZInteraction.wasKeyPressed('p')) { punchCounter = 0; hideHero(); heroPunch.show(); heroState = PUNCH; } if (jumpCounter > JUMPHEIGHT) { heroState = LAND; hideHero(); heroLand.show(); } else { posy -= 2; positionHero(posx, posy); if (EZInteraction.wasKeyPressed('h')) { heroState = HOVER; hideHero(); heroHover.show(); } } break; case PUNCH: // PUNCH punchCounter++; posx += 2; positionHero(posx, posy); if (punchCounter > PUNCHDISTANCE) { heroState = LAND; hideHero(); heroLand.show(); } break; case LAND: jumpCounter--;
- 4. if (jumpCounter <= 0) { heroState = STAND; hideHero(); heroStand.show(); } else { posy += 2; positionHero(posx, posy); if (EZInteraction.wasKeyPressed('h')) { heroState = HOVER; hideHero(); heroHover.show(); } } break; case HOVER: if (EZInteraction.wasKeyPressed('l')) { heroState = LAND; hideHero(); heroLand.show(); } break; } } } Solution public class Main2 { public static void main(String[] args) { EZ.initialize(512, 512); EZ.addImage("metropolis.jpg", 256, 256); // Hero myHero = new Hero(100,390);
- 5. HeroPlus mySecondHero = new HeroPlus(100,390); while(true){ //myHero.processStates(); mySecondHero.processStates(); EZ.refreshScreen(); } } } class HeroPlus { EZImage heroStand; EZImage heroJump; EZImage heroLand; EZImage heroHover; EZImage heroPunch; // PUNCH int posx = 0; int posy = 0; int heroState = STAND; static final int STAND = 1; static final int JUMP = 2; static final int LAND = 3; static final int HOVER = 4; static final int PUNCH = 5; static final int JUMPHEIGHT = 100; static final int PUNCHDISTANCE = 50; // PUNCH int punchCounter = 0; // PUNCH int jumpCounter = 0; HeroPlus(int x, int y) { posx = x; posy = y; heroStand = EZ.addImage("superstand.png", posx, posy);
- 6. heroLand = EZ.addImage("superland.png", posx, posy); heroJump = EZ.addImage("superjump.png", posx, posy); heroHover = EZ.addImage("superhover.png", posx, posy); heroPunch = EZ.addImage("superpunch.png", posx, posy); // PUNCH hideHero(); heroStand.show(); } void hideHero() { heroStand.hide(); heroLand.hide(); heroJump.hide(); heroHover.hide(); heroPunch.hide(); // PUNCH } void positionHero(int x, int y) { heroStand.translateTo(x, y); heroJump.translateTo(x, y); heroLand.translateTo(x, y); heroHover.translateTo(x, y); heroPunch.translateTo(x, y); // PUNCH } void processStates() { switch (heroState) { case STAND: if (EZInteraction.wasKeyPressed('j')) { heroState = JUMP; jumpCounter = 0; hideHero(); heroJump.show(); } break; case JUMP:
- 7. jumpCounter++; if (EZInteraction.wasKeyPressed('p')) { punchCounter = 0; hideHero(); heroPunch.show(); heroState = PUNCH; } if (jumpCounter > JUMPHEIGHT) { heroState = LAND; hideHero(); heroLand.show(); } else { posy -= 2; positionHero(posx, posy); if (EZInteraction.wasKeyPressed('h')) { heroState = HOVER; hideHero(); heroHover.show(); } } break; case PUNCH: // PUNCH punchCounter++; posx += 2; positionHero(posx, posy); if (punchCounter > PUNCHDISTANCE) { heroState = LAND; hideHero(); heroLand.show(); } break; case LAND: jumpCounter--; if (jumpCounter <= 0) { heroState = STAND;
- 8. hideHero(); heroStand.show(); } else { posy += 2; positionHero(posx, posy); if (EZInteraction.wasKeyPressed('h')) { heroState = HOVER; hideHero(); heroHover.show(); } } break; case HOVER: jumpCounter++; if (EZInteraction.wasKeyPressed('j')) { punchCounter = 0; hideHero(); heroPunch.show(); heroState = JUMP; } if (EZInteraction.wasKeyPressed('h')) { heroState = HOVER; hideHero(); heroHover.show(); } if (EZInteraction.wasKeyPressed('p')) { punchCounter = 0; hideHero(); heroPunch.show(); heroState = PUNCH; } if (EZInteraction.wasKeyPressed('l')) { heroState = LAND; hideHero(); heroLand.show(); }
- 9. break; } } }